New question on Pot Odds: Sklansky vs. Harrington

[Bill Kolter]

Quick quesiton on calculating pot odds. David Sklansky and Dan Harrington differ on how to calculate pot odds.

In "Theory of Poker," Sklansky includes the call bet in the pot odds numerator. In "Harrington on Hold 'em," Harrington does not.

Example.

$300 pot, $100 to call.

Sklansky: pot odds are $400/$100
Harrington: pot odds are $300/$100.

What is the accepted method of calculating pot odds?

[jojobinks]

seems odd. page #'s please?

[Bill Kolter]

Seemed odd to me, too. I don't have either book with me now, but if you have them, you might check the Sklansky index. As far as Harrington's book, he uses dozens of specific examples. Pretty much any page that has a problem on it (back of the chapters) will show how he calculates them.

Anyway, the question remains: what is the accepted method?

[Robk]

sklansky is correct. try the beginner forum for similar questions.

[David04]

[ QUOTE ]

Anyway, the question remains: what is the accepted method?

[/ QUOTE ]
Harrington's way.

[Geddon]

The "slansky method" is right (you stand to win $400 by betting $100) but are you sure Harrington does it the other way? I'm just starting his book but on page 92 he totals everyone's bets and calls into the pot for calculating odds.

[Robk]

i guess one of us is wrong. /images/graemlins/smile.gif i think the problem is in the way the poster stated his question.


[ QUOTE ]
$300 dollar pot, $100 to call

[/ QUOTE ]

i assumed this meant someone had bet $100 into a $300 pot. if the $300 dollar figure includes your opponents' $100 bet already im wrong.

[Leo Bello]

The correct method is to calculate:
The total pot to the moment including the bet u have to call (meaning the value from your opponent) : the amount u have to call.

On the flop, POT TOTAL 300.
Player A Bets: 100

Player B ODDS: 400 : 100 4:1

He must call 100 to win the 400 already in the pot.

[Bill Kolter]

I can certainly check it again, but I've read dozens of problems in the book and continually run into wording along these lines: (not a quote, just an example).

"If there is $300 in the pot, and it will cost you $100 to call, you stand to gain a $300 reward for a $100 bet, so you pot odds are 3-1." Note that he is talking about a player considering making a $100 bet into a pot that already has $300 into it. I interpret this as a $300 pot that will have $400 in it AFTER the call; hence, 4-1, as I have always understood it, as as Sklansky explains it.

Anyway, the consensus on this thread seems to be that the $100 you put in to call counts in the pot odds numerator, which is what I had always thought.

But I see your point: when a shlub like me asks the question, "Who's wrong: Sklansky or Harrington?" the correct answer MUST be: "It must be you."

Thanks to all for the input.

[stenaf]

That shouldn't be the general consensus, as it's wrong.

Please refer to a page in ToP where this is the case please. On page 36, this part of a sentance can be read "... you would be correct to call a $10 bet when the pot is $50 since your chanse of making the flush or the straight is better than 5-1".

You put $10 to win $50. Not $10 to win $50+$10. You don't win your own money, kind of.

[Geddon]

[ QUOTE ]
Anyway, the consensus on this thread seems to be that the $100 you put in to call counts in the pot odds numerator, which is what I had always thought.

[/ QUOTE ]

No, that's not right. You don't count YOUR $100 into the odds.

I thinkt he confusion comes from how the question is inturpreted. When I read "$300 in the pot and $100 to call", I took that to mean there wasn $300 IN THE MIDDLE and someone ELSE had bet $100, so the total amount you stand to win is $400.

[Bill Kolter]

This is exactly my question. I may be mis-understanding what I've read.

But if you adhere to the concept of "Once it's in the pot, it's not your money anymore," (Can't tell you how many times I've heard that), then 4-1 seems correct.

[Bluegoose75]

I guess it depends on if they are talking about odds or probabilities. If you ask the question about what is the probability that in a 52 card deck, completely randomly shuffled, what is the probability that a diamond will be the first card off the top then the probability is 13/52 or 1/4. However the odds against it happening are 3:1. So maybe (since I don't have either book handy) one of them is using probability language and the other is using odds against?

[Kevin K.]

It's been awhile since I've read Theory of Poker, but I don't recall pot odds being explained in this manner. If they are, I'm sure it was a typo/misprint, etc. Regardless, Harrington is correct.

300 in the pot and 100 to call equals 300-100 or 3-1 odds. By definition, pot odds are the ratio of the amount of money in the pot to the amount you must put in to continue.

Adding your call to the numerator is incorrect.

[Bill Kolter]

My apologies to all. At the risk of complicating what I thought was a fairly simple question, let me rephrase the situation as I understand it.

You're heads-up. The pot starts out at $200. First-to-act bets $100. There is now $300 in the pot. It's on you, $100 to call.

If you call, there will be $400 in the pot. Do you stand to win $400, or do you stand to win $300? Is the money you will bet money you will win, or does it not count?

Let's try a thought experiment... same situation: heads-up. It's $100 to you, and there is $300 in the pot. You raise it to $200. There is now $500 in the pot. Your opponent re-raises $200. There is now $700 in the pot, and it's $200 to call. Are the pot odds 7-2, 9-2, or are they only 5-2 because "you can't win back your original raise of $200?"

In other words, does your own bet affect pot odds? I have to conclude that it does, thus your bet should be included in the odds, whether you're about to make it or you have made it. Pot odds don't apply if you're not in the hand....correct? Thus, if you are considering what the odds are IF YOU PLAY, then it seems to me that the bet you are ABOUT to make must count. It will certainly count on the next betting round. The question, it seems to me, is not "What are the pot odds at this point in time before I bet," but "What are the pot odds vs. my chances of winning IF I BET?"

[wuwei]

[ QUOTE ]
This is exactly my question. I may be mis-understanding what I've read.

But if you adhere to the concept of "Once it's in the pot, it's not your money anymore," (Can't tell you how many times I've heard that), then 4-1 seems correct.

[/ QUOTE ]

No, it doesn't seem correct. It's not in the pot yet, so how can it impact the odds you're getting to call? Go read ToP p. 35 again. Then read the rest of the section. You're confused about how DS explains things. You seem to be thinking of a situation where you bet $100 into a pot that is already $300, which means you're offering your opponents 4:1 pot odds to call. If your opponent bets $100 and the pot is $300 *including* his $100 bet, you are getting 3:1 pot odds to call.

If you have a copy of SSH by Ed Miller, try looking at that instead. Ed writes: "If you must call a $4 bet, and the pot currently contains $40 (including the $4 bet), then your pot odds are 40:4 or 10:1."

[Bill Kolter]

Thanks for the clarification. I will re-read the Sklansky section. I appreciate your taking the time to look it up.

[Bill Kolter]

Could be. I'll re-read both and find out. I've probably just stirred up a whole lot of foam over a simple mis-understanding. Again, I apologize.

[Yosemite Mark]

[ QUOTE ]
...
If you call, there will be $400 in the pot. Do you stand to win $400, or do you stand to win $300? Is the money you will bet money you will win, or does it not count?


[/ QUOTE ]
Look at it this way: you are deciding if you should risk $100 to win the $300 in the pot. The $100 is not in the pot yet. That makes the odds 3:1.

[ QUOTE ]

Let's try a thought experiment... same situation: heads-up. It's $100 to you, and there is $300 in the pot. You raise it to $200. There is now $500 in the pot. Your opponent re-raises $200. There is now $700 in the pot, and it's $200 to call. Are the pot odds 7-2, 9-2, or are they only 5-2 because "you can't win back your original raise of $200?"


[/ QUOTE ]
Once the money is in the pot, no longer consider it yours. The only significant numbers are: What is in the pot now ($700 in the second part of your example), and what do you have to risk now to get a chance to win it (Here, $200). Thus the odds for your second example are 7:2.

[Bill Kolter]

Gotcha. All is clear.

Thanks! /images/graemlins/grin.gif

[wuwei]

[ QUOTE ]
Could be. I'll re-read both and find out. I've probably just stirred up a whole lot of foam over a simple mis-understanding. Again, I apologize.

[/ QUOTE ]

No problem, Bill. That's why the forums are here.

[Paul2432]

[ QUOTE ]
My apologies to all.

[/ QUOTE ]

No need to apoligize. The purpose of this forum is to discuss poker.

[ QUOTE ]
You're heads-up. The pot starts out at $200. First-to-act bets $100. There is now $300 in the pot. It's on you, $100 to call.

If you call, there will be $400 in the pot. Do you stand to win $400, or do you stand to win $300? Is the money you will bet money you will win, or does it not count?

[/ QUOTE ]

In this situation your pot odds are 3:1. Both Sklansky & Harrington would agree to this as would any experienced player in this forum.

[ QUOTE ]
Let's try a thought experiment... same situation: heads-up. It's $100 to you, and there is $300 in the pot. You raise it to $200. There is now $500 in the pot.

Your opponent re-raises $200. There is now $700 in the pot, and it's $200 to call. Are the pot odds 7-2, 9-2, or are they only 5-2 because "you can't win back your original raise of $200?"

[/ QUOTE ]

None of the above. Pot odds are now 8:2, because there is $800 in the pot not $700. The pot started at 200, opponent bet 100, you raised 100 to 200, and your opponent raised 200 to 400. The pot contains the original 200, 400 from your opponent and 200 from you. You need to put in 200 more to call.

I would like to point out one other problem from another of your posts. You wrote:

[ QUOTE ]
"If there is $300 in the pot, and it will cost you $100 to call, you stand to gain a $300 reward for a $100 bet, so you pot odds are 3-1." Note that he is talking about a player considering making a $100 bet into a pot that already has $300 into it. I interpret this as a $300 pot that will have $400 in it AFTER the call; hence, 4-1, as I have always understood it, as as Sklansky explains it.

[/ QUOTE ]

In this quote you have been loose with your terminology. There is a difference between making a bet and calling a bet. I think this may be the source of some of your confusion. If you make a bet where you want your opponent to fold your odds on the bet are (size of pot):(size of bet). Harrington talks a lot about continuation bets where you might bet half the pot giving yourself 2:1 on the bet. This is a completely different situation from your opponent betting half the pot, in which case you are getting odds of 3:1.

Paul

[Bill Kolter]

Thanks. Great response! This helps a lot. I think I get it now.

[jwombles]

[ QUOTE ]
My apologies to all. At the risk of complicating what I thought was a fairly simple question, let me rephrase the situation as I understand it.

You're heads-up. The pot starts out at $200. First-to-act bets $100. There is now $300 in the pot. It's on you, $100 to call.

If you call, there will be $400 in the pot. Do you stand to win $400, or do you stand to win $300? Is the money you will bet money you will win, or does it not count?

Let's try a thought experiment... same situation: heads-up. It's $100 to you, and there is $300 in the pot. You raise it to $200. There is now $500 in the pot. Your opponent re-raises $200. There is now $700 in the pot, and it's $200 to call. Are the pot odds 7-2, 9-2, or are they only 5-2 because "you can't win back your original raise of $200?"

In other words, does your own bet affect pot odds? I have to conclude that it does, thus your bet should be included in the odds, whether you're about to make it or you have made it. Pot odds don't apply if you're not in the hand....correct? Thus, if you are considering what the odds are IF YOU PLAY, then it seems to me that the bet you are ABOUT to make must count. It will certainly count on the next betting round. The question, it seems to me, is not "What are the pot odds at this point in time before I bet," but "What are the pot odds vs. my chances of winning IF I BET?"

[/ QUOTE ]

Whenever you are calculating your odds before you make a call, you don't include your bet. The reason is because your money or your call is not in the pot yet, ie, you can still fold and hold on to your 100.

For example, $300 in the pot and it's $100 to call means you WILL be getting 3 to 1 on your money. You don't include your $100 b/c it's NOT in the pot yet.

It's really that simple.

Much success,
Wombles

[starvs]

when using percentages to determine if you should call or not you DO include the money you will be putting into the pot to call in calculating what percent of the pot you are calling, correct?

[EarlCat]

[ QUOTE ]
when using percentages to determine if you should call or not you DO include the money you will be putting into the pot to call in calculating what percent of the pot you are calling, correct?

[/ QUOTE ]

I only think that matters if you are raising the pot...then you would (in your head) first call (adding to the pot) then raise the total of the pot after you called.

For instance, if there's $30 in the pot after the flop and your opponent bets $5, raising the pot would cost $45 ($5 call plus $40 raise).

[BugsBunny]

That's correct, because with percentages you're actually calculating equity. And your bet is part of the equity calculation.

Pot contains $400. Your hand has a 20% chance of winning. It's $100 to call.

This is a breakeven situation:
20% * (400 + 100) = 100

20% is the same as 4 to 1 ( (1 - .2)/.2 = 4 ) amd 4 to 1 is obviously break-even here.

[btwnfdngs]

For any newbies like me who have been misapplying this concept (what a leak!), I'd like to sort of belatedly restate what I think Bugs and others have said (thanks), because I'm not sure it's really been made explicit where the confusion may come from. I can't answer the original question of how to correctly define pot odds, and in fact I'm sure I'll bungle the terminology, but I want to point out how the distinction discussed in this thread affects the overall purpose of calculating pot odds, which of course is to be able to compare the pot odds "ratio" (I'll call it) to the chances of winning "ratio" to determine the correct play in the long run.

The most important thing is not to use the commonly accepted definition, but rather to use whatever definition you choose in conjunction with the appropriate corresponding method of expressing your chances of winning. To use Bugs' example, say the bet's to you, the pot is 400 before you bet, it's 100 to call, and your chances of winning the hand are 20 %. (Again, I know the terminology is wrong, but I hope the meaning is clear.) If you express pot odds as 4:1 (size of pot before you bet : your bet), then you must express your 20% chances of winning as 1:4 (number of times you'll win per : number of times you'll lose).

But if you do include your unmade bet the in pot odds ratio, so that your pot odds under this scenario would be expressed as 5 to 1, then the corresponding correct way to express your 20% chances of winning is, 1 in 5.

Obviously, if you were express your pot odds as 4:1 and your chances of winning as 1 in 5 (or 5 to 1 and 1:4), you wouldn't realize this is a break even bet (unless you've learned to make the translation in your head, which I'm sure many veterans do).

If, like me, you forget, think of a coin flip. If the prize for calling it correctly is 1 (the pot), and the bet is 1, then your thinking is either: 1 to 1 odds, 1:1 chances of winning, so it's break even; or 2 to 1 odds, 1 in 2 chances, so it's break even.

If any of this is wrong, please correct it and I'll never post again. If it's right, hopefully there's another newbie moron like myself out there who'll find it helpful.

[andyfox]

How can you win $400 if there's $300 in the pot? If the bet to you was $1,000,000, would you stand to win $1,000,300 if you call?

[Kevin K.]

[ QUOTE ]

But if you do include your unmade bet the in pot odds ratio, so that your pot odds under this scenario would be expressed as 5 to 1, then the corresponding correct way to express your 20% chances of winning is, 1 in 5.

Obviously, if you were express your pot odds as 4:1 and your chances of winning as 1 in 5 (or 5 to 1 and 1:4), you wouldn't realize this is a break even bet (unless you've learned to make the translation in your head, which I'm sure many veterans do).


[/ QUOTE ]

You're making this more complicated than it is. The problem is you can't include the unmade bet. Just don't do it. Anybody who participates in any form of gambling won't know what you're talking about when you call 5 to 1 and 1 in 5 the same thing, because they aren't. 5 to 1 = 1 in 6, period. You're speaking a different language altogether.

It seems that some of the confusion comes from attempting to convert the pot odds into a percentage. If that's the problem, just skip the conversion. It's not necessary. Just compare the pot odds to your odds to complete your hand, win the pot, etc.

Ex:
$90 in the pot, $10 to go = 9 to 1.
8 outs to hit your straight on the river = 38 to 8 = 4.75 to 1. Call.

You don't HAVE to know that you're 1 in 5.75 or 17% to make the call. The percentages, translating 9 to 1 to 1 in 10 or 10%, etc. will become second nature before too long.

[jimymat]

I dont remember the exact way Harrington worded it but I believe hes refering to the odds of your opponent folding if you bet out. Pot odds is pot odds . A lot of people responding to this post seem to have trouble undersrtanding how it works. If there is 200 in the pot ont thre flop and your opponent bets 100 there would be 300 in the pot. It is now costing you 100 to win 300 or 3-1 odds. If you are first to act instead, and there is 200 in the pot and you bet out 100 in hopes of your opponent folding, you have to have your opponent fold two thirds of the time for this to be a positive EV play. 2-1 equals 33% so if you think you can win the pot a third of the time and your opponent will fold 66% of the time , you win over time. This is used when you hold a weak hand and or are bluffing and your wondering how to calculate your odds of successfully bluffing.
Just off the top of my head Im assuming thats what Harrington is speaking of. Ive never seen Sk;ansky represent that figure anywhere. Do you have a book with page #? If I could see it I could elaborate more for you maybe.

[Shoe]

I agree... I think Sklansky and Harrington both know how pot odds are calculated. I think the confusion just comes from how the post was worded, and whether the opponent has already bet out or not.

[FatBob]

As I was reading the Harrington book, I thought there was a difference in pot odds calculations as well, but it was when someone raises behind you. I don't have my ToP book handy, so I could be remembering wrong, but lets take Harrington's problem hand 2-1 as an example:

You are player E

Both Blinds = 150
Player C bets 100
D folds
So there's 250 in the pot - giving you 2.5 to 1
You call for 100
The button (F)raises to 200
The blinds fold
Player C calls for the additional 100.

Harrington says the pot is 650 and your 100 needed to call gives you 6.5 to 1 odds.

If I'm not mistaken, Sklansky would say you have 5.5 to 2 odds here. You are putting in 200 to get the 550 that the other players put in.

If I'm recalling correctly, then this is a major difference. The Sklansky method is giving you less than half the odds Harrington is. (Of course its slightly better than the 2.5 to one that you had when you called the first time.)

[BugsBunny]

I think that the 5.5 to 2 figure would be Sklansky talking about effective odds as opposed to pot odds.

The pot odds on the 2nd call are indeed 6.5 to 1.
The effective odds that you got for the entire round are 5.5 to 2.